<span><span><span><span>2<span>y2</span></span>+y</span>−10</span>=0</span>Step 1: Factor left side of equation.<span><span><span>(<span><span>2y</span>+5</span>)</span><span>(<span>y−2</span>)</span></span>=0</span>Step 2: Set factors equal to 0.<span><span><span><span>2y</span>+5</span>=<span><span><span>0<span> or </span></span>y</span>−2</span></span>=0</span><span><span>y=<span><span><span><span>−5</span>2</span><span> or </span></span>y</span></span>=2</span>Answer:<span><span>y=<span><span><span><span>−5</span>2</span><span> or </span></span>y</span></span>=<span>2</span></span>
Step-by-step explanation:
volume of a cylinder is
ground area × height = pi×r²×h
r = 4
what we need to calculate is how high the water stands with one "serving" of 2m³.
that height is then added every minute (with the additional 2m³ during that minute).
so,
2 = pi×4²×h = pi×16×h
2/16 = 1/8 = pi×h
h = (1/8) / pi = 1/(8pi) = 0.039788736... m
that is rounded about 0.04 m or 4 cm.
therefore, the water rises by 0.04 m or 4 cm per minute
0.04m/min
Tonya's homework assignment has four problems a, b, c, d. Her teacher instructed her to simplify each expression and then find out its value. The value of a is 3/28, the value of b is 1/2, the value of c is 15/56 and the value of d is 1/15.
Each problem has two fractions. They are given below and should be simplified.
a) 2/7 × 3/8 = 1/7 × 3/4 = 3/28
b) 2/3 × 3/4 = 1/1 × 1/2 = 1/2
c) 3/7 × 5/8 = 15/56
d) 3/10 × 2/9 = 1/5 × 1/3 = 1/15
To know more about fractions:
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What we know:
Football field is 100 yards in length
End zones are 10 yards each in length
Perimeter between pylons is 306 2/3 yards
What we need to find:
a. Perimeter and area of one end zone
b. Perimeter and area of without end zones
c. Perimeter and area of playing field with end zones
First we need to find the measurements of the field between pylons using the perimeter of 306 2/3 yards. We already know the length is 100 yards so we need to find width (w).
P=2l + 2w
306 2/3=2 (100) + 2w
306 2/3= 200 + 2w
300 2/3-200=200-200+2w
106 2/3=2w
106 2/3/2=2/2w
53 1/3=w
a. Perimeter=2 (10)+2 (53 1/3)=126 2/3 yards
Area=10×53 1/3=533 1/3 yd²
b. Perimeter=2 (100)+2 (53 1/3)=306 2/3 yards
Area=100×53 1/3=5333 1/3 yd²
c. Perimeter=2 (120) + 2 (53 1/3)=346 2/3 yards
Area=120×53 1/3=6400 yd²
Answer: These Lines are <em>Parallel</em>
Step-by-step explanation: For two lines to be parallel, they must have the same slope with different y-intercepts. Both of the lines above have a slope of 3 and differing y-intercepts of 4 and -5. Therefore, these lines must be parallel to one another.
